(1)a^2=b^2=1,所以a^2-b^2=0,即(a-b)(a+b)=0,即a+b与a-b垂直.
(2)由x⊥y可知ka^2+3ab+ab+3b^2/k=0,化简得k+3/k+2sinα-2√3cosα=0,
即4sin(α-pi/3)=-(k+3/k)
(3)k∈R+,则=-(k+3/k)<=-2√3,即sin(α-pi/3)<=-√3/2
从而有4pi/3<=α-pi/3<=5pi/3,解之得5pi/3<=a<=2pi